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Consider the iterated function system $T_{1}(x)=(\beta x,\tau y)$, $T_{2}(x,y)=(\beta x+(1-\beta),\tau y+ (1-\tau))$ for $\beta\in(1/2,1)$ and $\tau\in (0,1/2)$ with self affine set $\Lambda_{\beta,\tau}$.

It is known that for almost all $\beta$ and all $\tau$ $\dim_{H}\Lambda_{\beta,\tau}=\dim b=1-\log(2\beta)/\log(\tau)$ where $b=(1/2,1/2)$ is the Bernoulli measure on $\Lambda$. Essential this is due to the fact that the projection of $b$ to the $x$-axis is an infinite Bernoulli convolution which is know to be generic absolutely continues (Solomyak theorem).

On the other hand we have shown that if $\beta^{-1}$ is a Pisot number we have $\dim b<\dim_{H}\Lambda_{\beta,\tau}<1-\log(2\beta)/\log(\tau)$ for all Bernoulli measures $b$. The projection of all $b$ in this case is singular with a dimension drop (Erdös theorem).

Now here comes the question: Is there a (ergodic) measure of full dimension and what is $\dim_{H}\Lambda_{\beta,\tau}$ in the later case?

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If $\beta^{-1}$ is Pisot, then there is an ergodic measure of maximal dimension. This is a special case of the rather difficult Theorem 2.15 in the paper Dimension Theory of iterated function systems by De-Jun Feng and Huyi Hu. Very roughly speaking, Feng and Hu adapt Ledrappier-Young theory to the IFS setting.

Note that the assumptions that $\beta^{-1}$ is Pisot and $\tau<1/2$ are crucial since they ensure that the weak separation condition holds (this is a key assumption in their theorem). As far as I know it is still not known if an ergodic measure of maximal dimension exists for all $\beta\in (1/2,1)$ (of course, this would follow from the above and the extremely difficult conjecture that the only singular Bernoulli convolutions come from Pisot numbers).

I don't know of any explicit formulas for the Hausdorff dimension of the attractor in the case that $\beta^{-1}$ is Pisot. In my paper Overlapping self-affine sets I gave a fairly explicit upper bound: $$ \dim_H(\Lambda_{\beta,\tau})\le 1-\frac{\log (2\beta)}{\log\tau} + \tau_\beta(q)-(1-q), $$ where $\tau_\beta$ is the (negative of the) $L^q$ spectrum of the (uniform) Bernoulli convolution of parameter $\beta$ and $q=\log\beta/\log\tau$ (See Theorem 15 and the remark afterward). It is well known that the multifractality of the BC for Pisot parameters ensures that $\tau_\beta(q)<1-q$.

I used to believe that this upper bound is in fact the Hausdorff dimension, but I don't have a proof.

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  • $\begingroup$ Dear Pablo, I am aware of the papers You cite. By the paper of De-Jun Feng and Huyi Hu we have exact dimensionality and a form of the Ledrappier Young formula for the dimension of ergodic measures on the attractor. I obtaion for the measure of maximal dimenison: $\dim_{H}\mu=1+\frac{h(\mu)-\log \beta^{-1}}{\log \tau^{-1}}$ where the entropy is bounded by $\log\beta_{1}^{-1}< h(\mu)<\log(2)$. But I do not see that this measure has full(!) Hasudorff dimenison. We have an upper bound on the Hausdorff dimension of the attractor but this does not meet my lower bound on $\dim_{H}\mu$. Best 9i $\endgroup$
    – Jörg Neunhäuserer
    Commented Sep 19, 2012 at 22:47
  • $\begingroup$ Dear Jörg, Feng and Hu do what you say but they also prove that there is in fact an ergodic measure of full dimension (i.e. the dimension of the attractor). It could be that my upper bound is not really sharp (through I don't see it immediately from your comment). $\endgroup$
    – Pablo Shmerkin
    Commented Sep 20, 2012 at 6:42
  • $\begingroup$ Dear Pablo, You are right, i have not realized it. Than $\dim_{H}\Lambda_{\beta,\tau}=\dim\mu$ where $\mu$ is the measure of maximal entropy for all measures with absolutely continuous projection (which is equivalent to full Garsia entropy). I still wonder if this quantity coincides with our upper bound on $\dim_{H}\Lambda_{\beta,\tau}$. If I look at lower bounds on $\dim\mu$ it does not seems so, but perhaps these bound are not sharp. $\endgroup$
    – Jörg Neunhäuserer
    Commented Sep 20, 2012 at 15:53
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Since no one has answered this after one week I'll give it a go, but I'm no expert in these things...

This question without the conditions on contraction coefficients was looked at in a paper of Pollicott and Weiss, Journal of Statistical Physics, 1994. They solve the question under various conditions - and the cases that they can't deal with appear to be hard. I'd look at their paper and other papers which cite it, but I'd expect any generalisation of their work to be beyond the scope of a mathoverflow question.

However, the fact that your second contraction coefficient is strictly less than 1/2 means that every point in your attractor has a unique 0-1 coding, so your system is isomorphic to the full shift on two symbols. You have a corresponding unique measure of maximal entropy (just push the 1/2-1/2 Bernoulli measure onto your system).

Often measures of maximal entropy correspond to measures of maximal dimension using a dimension=entropy/lyapunov exponents formula, I don't under what conditions this formula is valid but I'd imagine that it's standard and can be found in the book by Climenhaga and Pesin, see their lecture notes online if you don't have access to the book, http://www.math.uh.edu/~climenha/fractals.html .

As a final tip, to get the best advice you need to go to the right people. The kind of people that look at this kind of question are mainly ergodic theorists, I'm sure that if you'd tagged this question ergodic theory then Vaughn Climenhaga, Pablo Shmerkin and others would have found it and given you a much better answer than this one...

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  • $\begingroup$ Tom, for non-conformal constructions (such as the one in the question), the formula "dimension=entropy/Lyapunov exponent" breaks down, essentially because there are different Lyapunov exponents involved. There are generalizations (essentially versions of the Ledrappier-Young formula) invoving all Lyapunov exponents, and the formula tha Jörg states for almost every $\beta$ is a manifestation of that. But the Pisot case is exceptional and certainly much more difficult (there is an ergodic measure of maximal dim but it's not Bernoulli and certainly not (1/2,1/2)-Bernoulli). $\endgroup$
    – Pablo Shmerkin
    Commented Sep 19, 2012 at 10:00
  • $\begingroup$ Thanks Pablo, that clears up some things that I hadn't grasped. $\endgroup$
    – Tom Kempton
    Commented Sep 21, 2012 at 9:03

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